Ragib Posted December 24, 2006 Posted December 24, 2006 Lol no offence, but wth are you going on about...alot of this dusnt relate to anything... quantised relativity note that two frames disconnected or [math]\sqrt 2 [/math] form constant pythagoras constant, of 01.41 (pi= 03.141) that is 1.41 "X" plane, and 1.41 "Y" plane. A square root is two answers. Mathamically + and - , or geometrically 2dimensional plane, of x and y. That is the distance between two positions that are squared. Aka the distance corner to corner of a square is equal to length x 01.41 and that this constant applies to the fact that all physics are formed from constants, that expand from a minimum of two frames. This is drawn as a right trangle with equal length sides. and a hypotinuse = 01.41 * length C may infact expand out from this constant in some form or another but I have not yet looked into that. Where as pi is the constant that occurs for a given length arm rotated around a center point one full revolution. The distance the oustide of that length arm travels devided by the length of the arm * 2 = pi The reason for this is that a closed object like a trangle or a square, or a circle is a closed system, and an event in space is called a system. That event can only exist by a minimum of two frames which we call a closed system or apply to a shape. Shapes form from a closed system between frames. Wth is all of this? pi is approximately 3.14159265358979323846264338328, where as root 2 is aprrox 1.41421356237309504880168872421, the point is, why are they randomly put together...o wateva im not even guna bother
TriggerGrinn Posted December 25, 2006 Posted December 25, 2006 I understand your confusion. I can't elaborate on it too much right now. It simply a perspective of understanding fundamental basics. An event is change in time. It forms the universe. An event requries a minimum of two seperate frames. It can be force, motion, gravity, light etc, etc. two frames are a closed system. That is, the event that happens between them is happening to both of them at the same time, and it is inevitable. such as motion between two frames. A closed system aquires a measured constant. A geometrical perspective of this is a right triangle. The hypotinuse is the event. The two sides intersecting at the right angle are the square of the event, and the value of the event. such as force, energy. The right angled triangle has a constant relatoinship, in terms of length. that is, the sqrt of 2. how this applies is that there is an equal opposite value occuring for every 1 event. In the lorentz transformation for example; [math] \gamma =\frac {1}{ \sqrt{1- \frac{v^2}{c^2}}}[/math] the answer infact creates two equal but opposite gamma factors, that I understand is not considered. [math] \gamma (-) [/math] [math] \gamma (+)[/math] Or in otherwords -change in both frames -an equal value for each of the 2 sides of the right triangle (not the hypotenuse) -in which the hypotenuse is the event for each frame squared, [math]a^2 + b^2 = c^2[/math] That is, there is two frames, two hypenuses(events), two sides (normal two sides of a right triangle) One right triangle for each frame, but they share the same hypotenuse. This is a square, with two frames sharing one diagnal line through them, forming a one single event, for each frame. There is a point in which this shape can only come a point, or the universal constant C limit on change. Draw this in a 3 dimensional shape and you get a relationship that displays a simple geometric pentahedron.
TriggerGrinn Posted December 25, 2006 Posted December 25, 2006 I'm not running any posuedo science. I am in the beginning stages to formulating a geometry of space-time,
TriggerGrinn Posted December 26, 2006 Posted December 26, 2006 When we look at the lorentz, and the fact that we end up with two answers when using a square root. [math]\gamma (+)[/math] [math]\gamma (-)[/math] We see that we can re-express this equation as [math]\gamma = \frac {1}{\left (1- \frac {v^2}{c^2}\right )^{\,1/2}}[/math] and in doing so we arive at the proper expression of modeling the universe in mathamatics. We get an answer for 1 frame that is 1/2 of the whole system (1). The lorentz transformation as this; ..suggests there is null result for the one frame(in motion, due to the + and - result of a square root) or equal opposite dilations for each frame of the system[math]^b[/math]. However, the latter[math]^b[/math] does not appear to be the case. As we see the frame at rest[math]^a[/math] appears to remain constant, while the frame in motion[math]^b[/math] undergoes all the proposed dilations. Thus, the proper lorentz transformation should use the [math]^{\,1/2}[/math] that is, [math]\gamma = \frac {1}{\left (1- \frac {v^2}{c^2}\right )^{\,1/2}}[/math] with this we can see why [math]x^2[/math] [math]x^1[/math] [math]x^{\,0.5}[/math] or [math]x^{\,1/2}[/math] are the 3 functions of powers found in any one event. note [math]{\,1/2}[/math] is used to replace [math]\sqrt x [/math] For example; a position of a body in respect to an observer is written in this formulation. The observer is part of the system to develope the position. thus the position and the observer are infact two points on the graph. And through this system there is two sets of triangulation, in repsect to the position relative to each frame. As so we dicover each frame is 1/2 of the system. A simple explainaton of this; [math]+K_e = 1/2 \left (mv^2 \right )[/math] This is only 1/2 of the system in which momentum of each frame is involved in the interaction. So thus we do arrive with a negitive Ke for the object that is going to have 'work' applied to it [math]-K_e = 1/2 \left (mv^2 \right )[/math] Where we see, Kinetic energy total[math] K_{\,et} = mv^2[/math] The Kinetic energy total is of course a null result. That is the total energy does not change, but the state of energy of the system changes in and equal and opposite fashion. and as such supports discription of the universe being a closed system, where nothing is lost or destroyed, but only changes form.
Ragib Posted December 26, 2006 Posted December 26, 2006 I understand all of that, im quite acquainted with that, but my confusion was with why you put pi next to root 2..
TriggerGrinn Posted December 26, 2006 Posted December 26, 2006 I just wondered how and if they could be connected. I suppose I shouldnt of added any of that. I've had a rough week, and I appologize for posting while I can't articulate my thoughts very clearly. Take it easy.
TriggerGrinn Posted December 27, 2006 Posted December 27, 2006 nope. I mean take it easy as in, see ya later, take care.
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